Question

How can the following simple implementation of sum be faster?

private long sum( int [] a, int begin, int end ) {
    if( a == null   ) {
        return 0;
    }
    long r = 0;
    for( int i =  begin ; i < end ; i++ ) {
       r+= a[i];
    }
    return r;
}

EDIT

Background is in order.

Reading latest entry on coding horror, I came to this site: http://codility.com which has this interesting programming test.

Anyway, I got 60 out of 100 in my submission, and basically ( I think ) is because this implementation of sum, because those parts where I failed are the performance parts. I’m getting TIME_OUT_ERROR’s

So, I was wondering if an optimization in the algorithm is possible.

So, no built in functions or assembly would be allowed. This my be done in C, C++, C#, Java or pretty much in any other.

EDIT

As usual, mmyers was right. I did profile the code and I saw most of the time was spent on that function, but I didn’t understand why. So what I did was to throw away my implementation and start with a new one.

This time I’ve got an optimal solution [ according to San Jacinto O(n) -see comments to MSN below – ]

This time I’ve got 81% on Codility which I think is good enough. The problem is that I didn’t take the 30 mins. but around 2 hrs. but I guess that leaves me still as a good programmer, for I could work on the problem until I found an optimal solution:

Here’s my result.

I never understood what is those “combinations of…” nor how to test “extreme_first”

Answer 1

I don’t think your problem is with the function that’s summing the array, it’s probably that you’re summing the array WAY to frequently. If you simply sum the WHOLE array once, and then step through the array until you find the first equilibrium point you should decrease the execution time sufficiently.

int equi ( int[] A ) {
    int equi = -1;

    long lower = 0;
    long upper = 0;
    foreach (int i in A)
        upper += i;

    for (int i = 0; i < A.Length; i++)
    {
        upper -= A[i];
        if (upper == lower)
        {
            equi = i;
            break;
        }
        else
            lower += A[i];
    }

    return equi;
}